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Ask the Wizard #366

What is your recommended starting word in Wordle?

"Anonymous" .

To answer that, I first looked at the frequency of each letter in each position, based on the list of allowed Wordle solutions.

Letter Frequency in Wordle

Letter Pos. 1 Pos. 2 Pos. 3 Pos. 4 Pos. 5 Total
A 141 304 307 163 64 979
B 173 16 57 24 11 281
C 198 40 56 152 31 477
D 111 20 75 69 118 393
E 72 242 177 318 424 1233
F 136 8 25 35 26 230
G 115 12 67 76 41 311
H 69 144 9 28 139 389
I 34 202 266 158 11 671
J 20 2 3 2 0 27
K 20 10 12 55 113 210
L 88 201 112 162 156 719
M 107 38 61 68 42 316
N 37 87 139 182 130 575
O 41 279 244 132 58 754
P 142 61 58 50 56 367
Q 23 5 1 0 0 29
R 105 267 163 152 212 899
S 366 16 80 171 36 669
T 149 77 111 139 253 729
U 33 186 165 82 1 467
V 43 15 49 46 0 153
W 83 44 26 25 17 195
X 0 14 12 3 8 37
Y 6 23 29 3 364 425
Z 3 2 11 20 4 40

Then I looked at all the words in the Wordle solution list with five distinct letters and scored them according to the letter frequency table above. I awarded two points for a match in the correct position and one point for a match in an incorrect position. Then I sorted the list, which you see below.

Best Starting Words in Wordle

Rank Word Points
1 Stare 5835
2 Arose 5781
3 Slate 5766
4 Raise 5721
5 Arise 5720
6 Saner 5694
7 Snare 5691
8 Irate 5682
9 Stale 5665
10 Crate 5652
11 Trace 5616
12 Later 5592
13 Share 5562
14 Store 5547
15 Scare 5546
16 Alter 5542
17 Crane 5541
18 Alert 5483
19 Teary 5479
20 Saute 5475
21 Cater 5460
22 Spare 5457
23 Alone 5452
24 Trade 5449
25 Snore 5403
26 Grate 5403
27 Shale 5392
28 Least 5390
29 Stole 5377
30 Scale 5376
31 React 5376
32 Blare 5368
33 Parse 5351
34 Glare 5340
35 Atone 5338
36 Learn 5324
37 Early 5320
38 Leant 5307
39 Paler 5285
40 Flare 5280
41 Aisle 5280
42 Shore 5274
43 Steal 5268
44 Trice 5267
45 Score 5258
46 Clear 5258
47 Crone 5253
48 Stone 5253
49 Heart 5252
50 Loser 5251
51 Taper 5248
52 Hater 5243
53 Relay 5241
54 Plate 5240
55 Adore 5239
56 Sauce 5236
57 Safer 5235
58 Alien 5233
59 Caste 5232
60 Shear 5231
61 Baler 5230
62 Siren 5226
63 Canoe 5215
64 Shire 5213
65 Renal 5210
66 Layer 5206
67 Tamer 5200
68 Large 5196
69 Pearl 5196
70 Route 5194
71 Brace 5192
72 Slice 5178
73 Stage 5171
74 Prose 5170
75 Spore 5169
76 Rouse 5166
77 Grace 5164
78 Solar 5152
79 Suite 5150
80 Roast 5145
81 Lager 5130
82 Plane 5129
83 Cleat 5129
84 Dealt 5128
85 Spear 5126
86 Great 5126
87 Aider 5123
88 Trope 5116
89 Spire 5108
90 Tread 5107
91 Slave 5097
92 Close 5090
93 Lance 5090
94 Rinse 5088
95 Cause 5087
96 Prone 5087
97 Drone 5082
98 Noise 5079
99 Crest 5073
100 Sober 5068

So, there you have it, my recommended starting word, which I use, is STARE.

What is i^i

Aidan

e^(-pi/2) =~ 0.20788.

 

Here is my solution (PDF).

Suppose a casino has a game based on a fair coin flip that pays even money. A player wishes to play one million times at $1 a bet. How much money should he bring to the table to have a 50% chance of not going broke?

Ace2

Let's first answer the question of what is the probability the player will be down more than x units after one million flips, assuming the player has an unlimited bankroll.

Since this is a fair bet, the mean win after a million flips is zero. The variance of each flip is 1, so the variance of one million flips is one million. One standard deviation is thus sqrt(1,000,000) = 1000.

We can find the bankroll required with the Excel function =norm.inv(probability,mean,standard deviation). For example, if we put in =norm.inv(.25,0,1000), we get -674.49. This means if after one million flips, the player has a 25% chance of being down 674 or more. Please keep in mind this is an estimate. To get a true answer, we should use the binomial distribution, which would be very tedious with a million flips.

It could very well happen that if the player took $674 to the table, he might run out of money before the million flips. If he could keep playing on credit, it might happen that he has a recovery and finishes less than $674 down. In fact, once the player is at -674, there is a 50/50 chance he will end up above or below -674 at any given point in the future.

So, if the player can play on credit, there are three possible outcomes.

  1. Player never falls below -674.

  2. Player falls below -674 at some point, but recovers and finishes above -674.
  3. Player falls below -674 at some point, keeps playing and loses even more.

We have established scenario 3 has a probability of 25%.

Scenario 2 must have the same probability as scenario 3, because once the player is down -674, he has a 50/50 chance to finish above or below that point after one million flips.

Scenario 1 is the only other alternative, which must have probability 100%-25%-25% = 50%.

If the probability the player never falls below 674 is 50%, then the alternative of falling below must be 100%-50% = 50%.

So, there is our answer to the original question, $674.

This question is asked and discussed in my forum at Wizard of Vegas.

You wish to play a game that requires two ordinary six-sided dice. Unfortunately, you lost the dice. However, you have nine index cards, which you may mark any way you like. The player must choose two index cards randomly from the nine, without replacement, and take the sum of the two cards.

Gialmere

The card values do not have to be integers.

 

Mark the cards as follows:

1 @ 0.5
1 @ 1.5
2 @ 2.5
1 @ 3.5
2 @ 4.5
1 @ 5.5
1 @ 6.5

This question is asked and discussed in my forum at Wizard of Vegas.